I disagree with your answer for c. The probability of exactly one 0 is 0.2916. However, given that there is exactly one 0, there are only 9 possibilities for the remaining 4 digits. Therefore, the probability of exactly one 7 is 4*(1/9)*(8/9)^3=2048/6561. Then, the probability of exactly one 0 and one 7 is (729/2500)*(2048/6561)=512/5625. For n digits, the probability is (n-1)/10*(9/10)^(n-2)*(n-1)/9*(8/9)^(n-2)=((n-1)^2)/90*(4/5)^(n-2).